Potential Energy of a system of Point Charges

[Tybstar]

Homework Statement

Two identical charges q are placed on the x axis, one at the origin and the other at x = 5 cm. A third charge -q is placed on the x-axis so the potential energy of the three-charge system is the same as the potential energy at infinite separation. Its x coordinate is: (x=13)

Homework Equations

I know the correct answer to this one because I'm looking over a marked up test in preparation for the final.

I've been using the equations for the electric potential energy of a system of point charges:

$$U=W=k\frac{q_{1}q_{2}}{r}$$

The Attempt at a Solution

...and expanding that to the work of the entire system:

$$W=0=W_{12}+W_{23}+W_{13}$$

so,$$W=0=k\frac{q_{1}q_{2}}{5}+k\frac{q_{2}q_{3}}{d}+k\frac{q_{1}q_{3}}{5+d}$$

Since $$q_{3}$$ is negative, simplify:

$$W=0=k\frac{q^{2}}{5}-k\frac{q^{2}}{d}-k\frac{q^{2}}{5+d}$$I might be missing something simple in algebra, but I can't figure out how to simplify this down enough to solve for $$d$$.

Any ideas? Am I even on the right track here?

Thanks!

 

[chaoseverlasting]

$$kq^2$$ can be eliminated (as the right hand side is equal to zero), and this leaves you with:

$$\frac{1}{5} -\frac{1}{d} -\frac{1}{5+d}=0$$. Take the negative terms to one side and cross multiply to get a quadratic equation in d. Solve quadratic for d. There are two possible solutions as this is a quadratic.

 

[Tybstar]

I'm sorry, I forgot to mention that I tried doing a quadratic, but was unable to get the correct answer. Also, I think that the correct term should be $$\frac{1}{(d-5)}$$, not $$\frac{1}{5+d}$$.

 

[chaoseverlasting]

Yes. You're right. I didnt look at the whole post. The solution, however, is $$x=\frac{15}{2}+-\frac{5\sqrt{5}}{2}$$